std :: aligned_storage的目的是什么？ [英] What is the purpose of std::aligned_storage?

问题描述

如果我理解正确的话， std :: aligned_storage 的主要优点是它可以管理对齐方式。也可以使用 memcpy（）复制它，并可以与POD类型一起使用。

If I understand it correctly, the main advantage of the std::aligned_storage is that it manages the alignment. Also it can be copied with memcpy() and can be used with POD types.

但是！

1）默认情况下，编译器会对齐POD类型，我们可以使用 #pragma pack（push，1）覆盖此编译器的对齐方式

1) The POD types are aligned by compiler by default and we can override this compiler's alignment using #pragma pack(push, 1)

2）默认情况下，我们可以使用 memcpy（）复制POD（我们不应该

2) We can copy POD with memcpy() by default (we shouldn't do something for this ability)

所以我真的不明白为什么我们需要 std :: aligned_storage ？

So I don't really understand why do we need std::aligned_storage?

推荐答案

只要想取消耦合，就可以使用 std :: aligned_storage

You can use std::aligned_storage whenever you wish to decouple memory allocation from object creation.

您声称：

也可以使用

Also it is usable only with POD types.

但这不是事实。没有什么能阻止 std :: aligned_storage 与非POD类型一起使用。

But this is not true. There is nothing preventing std::aligned_storage from being used with non-POD types.

关于cppreference的示例提供了一个合法的用例：

The example on cppreference provides a legitimate use case:

template<class T, std::size_t N>
class static_vector
{
    // properly aligned uninitialized storage for N T's
    typename std::aligned_storage<sizeof(T), alignof(T)>::type data[N];
    std::size_t m_size = 0;
...

这里的想法是构造 static_vector ，立即为 T N 个对象分配内存。 c $ c>，但尚未创建类型为 T 的对象。

The idea here is that once the static_vector is constructed, memory is immediately allocated for N objects of type T, but no objects of type T are created yet.

您无法使用简单的方法 T data [N]; 数组成员，因为它将立即为每个元素运行 T 的构造函数，否则即使 T 不是默认可构造的，甚至都不会编译。

You cannot do that with a simple T data[N]; array member, because this would immediately run T's constructor for each element, or wouldn't even compile if T is not default-constructible.

这篇关于std :: aligned_storage的目的是什么？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！